[LC] 144. Binary Tree Preorder Traversal】的更多相关文章

Given a binary tree, return the preorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,2,3] Solution 1: # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = N…
144. Binary Tree Preorder Traversal Difficulty: Medium Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it it…
144. Binary Tree Preorder Traversal 前序的非递归遍历:用堆来实现 如果把这个代码改成先向堆存储左节点再存储右节点,就变成了每一行从右向左打印 如果用队列替代堆,并且先存储左节点,再存储右节点,就变成了逐行打印 class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> result; if(root == NULL) return res…
Binary Tree Preorder Traversal Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 解法一:递归 /** *…
Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 树的遍历,最常见的有先序遍历,中序遍历,后序遍历和层序遍历,它们用递归实现起来都非常的简…
Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. 二叉树的前序遍历,根节点→左子树→右子树 解题思路一: 递归实现,JAVA实现如下: public List<Integer> preorderTraversal(TreeNode root) { List<I…
Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. 代码如下: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNod…
Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 求前序遍历,要求不用递归.     使用双向队列.   /** * Definition…
题目描述: Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. 解题思路: 这个问题最简单的方法是使用递归,但是题目规定不能使用,得使用迭代的方法. 那么我们考虑使用栈来实现. 思路是每次遍历这节点,把该点的值放入list中,然后把该点的右孩子放入栈中,并将当前点设置为左…
题目: Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 提示: 首先需要明确前序遍历的顺序,即:节点 -> 左孩子 -> 右孩子,这一…
Given a binary tree, return the preorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,2,3] Follow up: Recursive solution is trivial, could you do it iteratively? ---------------------------------------------------…
Given a binary tree, return the preorder traversal of its nodes' values. Example: Input: [,,] \ / Output: [,,] Follow up: Recursive solution is trivial, could you do it iteratively? 方法一:使用迭代(C++) vector<int> preorderTraversal(TreeNode* root) { vecto…
Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? Subscribe to see which companies asked this…
Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? class Solution { public: vector<int> preor…
Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree [1,null,2,3], 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 递归 : class Solution { private List<Intege…
作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 递归 迭代 日期 题目地址:https://leetcode.com/problems/binary-tree-preorder-traversal/#/description 题目描述 Given a binary tree, return the preorder traversal of its nodes' values. For exampl…
题目意思:二叉树先序遍历,结果存在vector<int>中 解题思路:1.递归(题目中说用递归做没什么意义,我也就贴贴代码吧) 2.迭代 迭代实现: class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> ans; if(root){ TreeNode* temp; stack<TreeNode*> s; //利用栈,每次打印栈顶,然后将栈顶弹出…
前序遍历:根左右 //用栈来实现非递归解法/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<int> preorderTraversal(T…
给定一棵二叉树,返回其节点值的前序遍历.例如:给定二叉树[1,null,2,3],   1    \     2    /   3返回 [1,2,3].注意: 递归方法很简单,你可以使用迭代方法来解决吗?详见:https://leetcode.com/problems/binary-tree-preorder-traversal/description/ Java实现: 递归实现: /** * Definition for a binary tree node. * public class T…
二叉树的基础操作:二叉树的先序遍历(详细请看数据结构和算法,任意本书都有介绍),即根,左子树,右子树,实现方法中还有用栈实现的,这里不介绍了 /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution…
参考例子:[8,3,1,6,4,7,10,14,13] 8,3,1 和 6,4 说明从root开始,沿着左臂向下寻找leaf 的过程中应该逐个将node.val push入ans. class Solution(object): def preorderTraversal(self, root): """ :type root: TreeNode :rtype: List[int] """ stack = [] ans = [] while st…
先序遍历的非递归办法,还是要用到一个stack class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> ret; if(!root) return ret; stack<TreeNode*> stk; stk.push(root); //ahd(root) //a(stk) //a(ret) ){ TreeNode* cur=stk.top(); stk.p…
关于二叉树的遍历请看: http://www.cnblogs.com/stAr-1/p/7058262.html /* 考察基本功的一道题,迭代实现二叉树前序遍历 */ public List<Integer> preorderTraversal(TreeNode root) { List<Integer> res = new ArrayList<>(); if (root==null) return res; Stack<TreeNode> stack =…
要求 二叉树的前序遍历 实现 递归 栈模拟          定义结构体 Command 模拟指令,字符串s描述命令,树节点node为指令作用的节点 定义栈 Stack 存储命令 1 #include <iostream> 2 #include <vector> 3 #include <stack> 4 #include <cassert> 5 6 using namespace std; 7 8 struct TreeNode{ 9 int val; 10…
144. 二叉树的前序遍历 144. Binary Tree Preorder Traversal 题目描述 给定一个二叉树,返回它的 前序 遍历. LeetCode144. Binary Tree Preorder Traversal 示例: 输入: [1,null,2,3] 1 \ 2 / 3 输出: [1,2,3] 进阶: 递归算法很简单,你可以通过迭代算法完成吗? Java 实现 Iterative Solution import java.util.LinkedList; import…
Binary Tree Preorder Traversal Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? Solution: 递归…
详见:剑指 Offer 题目汇总索引:第6题 Binary Tree Postorder Traversal            Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note: Recursive solution is trivial, could…
原题 Binary Tree Preorder Traversal 没什么好说的... 二叉树的前序遍历,当然如果我一样忘记了什么是前序遍历的..  啊啊.. 总之,前序.中序.后序,是按照根的位置来决定的,根首先访问,是前序. /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(N…
二叉树的非递归前序遍历,大抵是很多人信手拈来.不屑一顾的题目罢.然而因为本人记性不好.基础太差的缘故,做这道题的时候居然自己琢磨出了一种解法,虽然谈不上创新,但简单一搜也未发现雷同,权且记录,希望于人于己皆有帮助. 估计能看到这里的读者,都是见过题目的,但还是链接原题在此:Binary Tree Preorder Traversal 二话不说,直接上代码: /** * Definition for binary tree * public class TreeNode { * int val;…
Binary Tree Preorder Traversal Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. c++版: /** * Definition for binary tree * struct TreeNode { * int val; * TreeNod…